cf-723C Polycarp at the Radio

cf-723C Polycarp at the Radio 

題目點擊打開鏈接  http://codeforces.com/contest/723/problem/C

分析：先定義每個數字在數組中出現次數叫做這個數的頻率，注意是次數。

題意：給出兩個數字 n 和 m，再給出 n 個數字放在數組 a[i] 裏。使數組 a 裏面的數全部變爲 1~m 之間的數字，並且使每個數的頻率儘量大，也就是每個數的頻率高於平均值。然後輸出數組改變後所有數頻率中的最小值 和 改變數組時改變了幾次，以及改變後的數組。

可以知道平均值就是 n/m，然後在數組a中找到符合 1~m 的數字並記錄下次數，如果高於平均值就再找下一個，低於平均值則記下差幾個數能夠到達平均值，這樣可以求出改變次數。最後在數組中找到所有不符合條件的數【頻率太小需要將頻率大的改爲這個數，大於m的需要變小】改掉

Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented as a sequence a₁, a₂, ..., a_n, where a_i is a band, which performs the i-th song. Polycarp likes bands with the numbers from 1 to m, but he doesn't really like others.

We define as b_j the number of songs the group j is going to perform tomorrow. Polycarp wants to change the playlist in such a way that the minimum among the numbers b₁, b₂, ..., b_m will be as large as possible.

Find this maximum possible value of the minimum among the b_j (1 ≤ j ≤ m), and the minimum number of changes in the playlist Polycarp needs to make to achieve it. One change in the playlist is a replacement of the performer of the i-th song with any other group.

Input

The first line of the input contains two integers n and m (1 ≤ m ≤ n ≤ 2000).

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹), where a_i is the performer of the i-th song.

Output

In the first line print two integers: the maximum possible value of the minimum among the b_j (1 ≤ j ≤ m), where b_j is the number of songs in the changed playlist performed by the j-th band, and the minimum number of changes in the playlist Polycarp needs to make.

In the second line print the changed playlist.

If there are multiple answers, print any of them.

Example

Input

4 2
1 2 3 2

Output

2 1
1 2 1 2 

Input

7 3
1 3 2 2 2 2 1

Output

2 1
1 3 3 2 2 2 1 

Input

4 4
1000000000 100 7 1000000000

Output

1 4
1 2 3 4 

Note

In the first sample, after Polycarp's changes the first band performs two songs (b₁ = 2), and the second band also performs two songs (b₂ = 2). Thus, the minimum of these values equals to 2. It is impossible to achieve a higher minimum value by any changes in the playlist.

In the second sample, after Polycarp's changes the first band performs two songs (b₁ = 2), the second band performs three songs (b₂ = 3), and the third band also performs two songs (b₃ = 2). Thus, the best minimum value is 2.

#include<iostream>
#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2050;
int a[N],val[N],p[N];
int pos[2005];
int main(){
	int n,m,min_ave,min_cnt;
	memset(pos,0,sizeof(pos));
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	min_ave=n/m;
	min_cnt=0;
	for(int i=1;i<=m;i++){
		int num=0;
		for(int j=1;j<=n;j++){
			if(a[j]==i){
				pos[j]=1;
				num++;
			}
			if(num>=min_ave){
				break;
			}
		}
		if(num<min_ave){
			val[i]=min_ave-num;
			min_cnt+=val[i];
		}
	}
	int count=1;
	for(int i=1;i<=m;i++){
		for(int j=0;j<val[i];j++){
			while(pos[count]==1){
				count++;
			}
			a[count++]=i;
		}
	}
	cout<<min_ave<<" "<<min_cnt<<endl;
	for(int i=1;i<=n;i++){
		cout<<a[i]<<" ";
	}
	cout<<endl;
	return 0;
}